Let $\alpha$ be the $2$-form $x^{2p-1} y^{-p-1} dx \wedge dy$ and consider the automorphism $\phi(x,y) = (x+y,y)$ of $X$. This preserves the divisor $D$ and acts trivially on $D$, so $\alpha$ and $\phi^{\ast}(\alpha)$ should have the same residue. In other words, $\phi^{\ast} \alpha - \alpha$ should have residue $0$. Now, $\phi^{\ast} \alpha - \alpha = \sum_{j=0}^{2p-2} \binom{2p-1}{j} x^j y^{p-2-j} dx \wedge dy$. The term $x^j y^{p-2-j} dx \wedge dy$ is exact whenever $j \not \equiv -1 \bmod p$, so (if we want a map that factors through $H^n(U)$), the residue of $\binom{2p-1}{p-1} x^{p-1} y^{-1} dx \wedge dy$ should be $0$. Also, $\binom{2p-1}{p-1} \equiv 2 \bmod p$ by Lucas' theorem. So the residue of $x^{p-1} y^{-1} dx \wedge dy$ should be $0$ in $H_{DR}^1(D)$. Note that $x^{p-1} dx$ is not zero in $H_{DR}^1(D)$, so we have to do something non-obvious.

I went looking for something more elementary to do and came up with an interesting idea: Although $x^{p-1} dx$ isn't exact, it is in a sense "almost exact". The exact forms are the kernel of the Cartier operator, and $x^{p-1} dx$ is in the kernel of the square of the Cartier operator. Define $EH^n$ to be $n$-forms modulo forms which are killed by some power of the Cartier operator. (This is a definition for top dimensional forms only; see my recent question for what I think the more general definition should be.) I think I can build a Gysin map $EH^n(X) \to EH^{n-1}(D)$. But I'm going to wait a bit to see if someone answers my other question before I write more.

Let $\alpha$ be the $2$-form $x^{2p-1} y^{-p-1} dx \wedge dy$ and consider the automorphism $\phi(x,y) = (x+y,y)$ of $X$. This preserves the divisor $D$ and acts trivially on $D$, so $\alpha$ and $\phi^{\ast}(\alpha)$ should have the same residue. In other words, $\phi^{\ast} \alpha - \alpha$ should have residue $0$. Now, $\phi^{\ast} \alpha - \alpha = \sum_{j=0}^{2p-2} \binom{2p-1}{j} x^j y^{p-2-j} dx \wedge dy$. If we compute residues naively, the residue of $\sum_{j=0}^{2p-2} \binom{2p-1}{j} x^j y^{p-2-j} dx \wedge dy$ should be $\binom{2p-1}{p-1} x^{p-1} dx$. Also, $\binom{2p-1}{p-1} \equiv 2 \neq 0 \bmod p$ by Lucas' theorem. But $x^{p-1} dx$ is not exact in characteristic $p$. So working naively can't give us a residue which is well defined in $H^{\ast}_{dR}$. Moreover, it doesn't make sense to fix this by defining $\binom{2p-j}{p-j} x^j y^{p-2-j} dx dy$ to have a nonzero residue for other values of $j$, because $x^j y^{p-2-j} dx dy$ is exact for all $0 \leq j \leq 2p-2$ except $p-1$.

I went looking for something more elementary to do and came up with an interesting idea: Although $x^{p-1} dx$ isn't exact, it is in a sense "almost exact". The exact forms are the kernel of the Cartier operator, and $x^{p-1} dx$ is in the kernel of the square of the Cartier operator. Define $EH^n$ to be $n$-forms modulo forms which are killed by some power of the Cartier operator. (This is a definition for top dimensional forms only; see my recent question for what I think the more general definition should be.) I think I can build a Gysin map $EH^n(X) \to EH^{n-1}(D)$. But I'm going to wait a bit to see if someone answers my other question before I write more.

Okay, let me spell out this idea in a bit more detail.

First of all, let's recall how residue works when $\omega$ only has a simple pole along $D$. First, choose an open set $X'$ on which $D$ is principal, with generator $t$, and on which there is a vector field $\vec{v}$ with $\langle \vec{v}, dt \rangle = 1$. Set $U' = X' \cap U$ and $D' = X \cap D$. If $\omega$ has only a simple pole on $D'$, then $t \omega$ extends to $X'$. Contracting $t \omega$ against $\vec{v}$ gives an $(n-1)$-form, which we can then restrict to $D'$. The final result is independent of the choices of $t$ and $\vec{v}$, and is the residue of $\omega$ to $D'$. We can cover $X$ by open sets $X'$ as above and compute the residue on each such set, and since the result is independent of our choices, we get a well defined residue on $D$. Nothing here uses characteristic $0$ (and we even get a specific differential form for our residue, not a cohomology class.)

Now, suppose that $\omega$ has a pole of order $N$, and let $\mathcal{C}$ be the Cartier operator. Then $\mathcal{C}(\omega)$ has a pole of order at most $1+(N-1)/p$. Applying the Cartier operator $k$ times for $k$ large enough that $p^k \geq N$, we get a differential form with a pole of order $\leq 1$. We can take the residue $\mathrm{Res}(\mathcal{C}^k \omega)$ of that form. But then we should apply the "inverse Cartier operator" $k$-times to this residue. The Cartier operator from top dimensional forms to top dimensional forms is surjective, but has a kernel, so what this really means is to find some $n-1$ form $\alpha$ on $D$ with $\mathcal{C}^k(\alpha) = \mathrm{Res}(\mathcal{C}^k \omega)$. So $\alpha$ is only defined modulo the kernel of $\mathcal{C}^k$. In other words, this residue is a class in $EH^{n-1}(D)$ in the sense I describe above. This is a map $\Omega^n(U) \to EH^{n-1}(D)$. It is also not hard to show that this map passes down to a map $EH^n(U) \to EH^{n-1}(D)$.

I don't know if this is helpful, but I think it is the best you can do.